Tetration And Telescoping

Let

$f\left( x \right) =\prod _{ k=0 }^{ n }{ \left( 1+\frac { 1 }{ { x }^{ { 2 }^{ k } } } \right) } \\ \prod _{ k=0 }^{ n }{ \left( 1+\frac { 1 }{ { x }^{ { 2 }^{ k } } } \right) } =\frac { \prod { 1+{ x }^{ { 2 }^{ k } } } }{ x^{ \sum { { 2 }^{ k } } } } = \frac { \prod { 1-x^{ { 2 }^{ k+1 } } } }{ { x }^{ { 2 }^{ n+1 }-1 }\prod { 1-{ x }^{ { 2 }^{ k } } } } =\frac { x(1-{ x }^{ { 2 }^{ n+1) } }) }{ { x }^{ { 2 }^{ n+1 } }(1-x) }$

Now

$\ln { f\left( x \right) } =\sum _{ k=0 }^{ n }{ \ln { \frac { { x }^{ { 2 }^{ k } }+1 }{ { x }^{ { 2 }^{ k } } } } } \Rightarrow \frac { f'\left( x \right) }{ f\left( x \right) } =\frac { -\sum _{ k=0 }^{ \infty }{ \frac { { 2 }^{ k } }{ 1+{ x }^{ { 2 }^{ k } } } } }{ x } \\ \Rightarrow { S }_{ n }=\sum _{ k=0 }^{ n }{ \frac { { 2 }^{ k } }{ 1+{ x }^{ { 2 }^{ k } } } } = \frac { -xf'\left( x \right) }{ f\left( x \right) }$

If $\left| x \right| >1$ , $\displaystyle \lim _{ n \rightarrow \infty }{ f\left( x \right) = \frac { x }{ x-1 } }$ and $\displaystyle \lim _{ n \rightarrow \infty }{ f'\left( x \right) = \frac { -1 }{ { (x-1) }^{ 2 } } }$

Therefore

$\lim _{ n \Rightarrow \infty }{ { S }_{ n } } = \boxed { \frac { 1 }{ x-1 } } \\ \therefore \omega =\sum _{ x=2 }^{ \infty }{ \sum _{ k=0 }^{ \infty }{ \frac { { 2 }^{ k } }{ 1+{ x }^{ { 2 }^{ k } } } { (-1) }^{ x } } } =\sum _{ x=2 }^{ \infty }{ \frac { 1 }{ x-1 } } { (-1) }^{ x }= \boxed { \ln { 2 } } \approx 0.6931\\ \therefore \left\lfloor 1000\omega \right\rfloor = \boxed { 693 }$

Another solution:

First observe that, it is not difficult to prove that $\sum_{n\ge 0}\frac{2^n}{1+z^{2^n}}=\frac{1}{z-1}$ $\forall z\in \mathbb{C}$ , such that $|z|>1$ . Thus the given sum is $S=\sum_{x\ge 2}\frac{(-1)^x}{x-1}=1-\frac{1}{2}+\frac{1}{3}+\cdots=\ln 2$ which gives the answer ${\lfloor1000\ln 2\rfloor}=\boxed{693}$

$\begin{aligned} &\large \sum_{x=2}^\infty \sum_{n=0}^\infty \dfrac{2^n}{1 + x^{2^n}} (-1)^x = \large \sum_{x=2}^\infty (-1)^x \Bigg(\sum_{n=0}^\infty \dfrac{2^n}{1 + x^{2^n}}\Bigg) \\ &=\large \sum_{x=2}^\infty (-1)^x \Bigg( \lim_{n\to\infty}\bigg(\dfrac{2^0}{1+x^{2^0}} + \dfrac{2^1}{1+x^{2^1}} + \dfrac{2^2}{1+x^{2^2}}+ \cdots+ \dfrac{2^n}{1+x^{2^n}}\bigg)\Bigg) \\ &=\large \sum_{x=2}^\infty (-1)^x \Bigg( \lim_{n\to\infty}\bigg( -\dfrac{1}{1-x}+\bigg(\dfrac{2^0}{1-x^{2^0}}+\dfrac{2^0}{1+x^{2^0}}\bigg) + \dfrac{2^1}{1+x^{2^1}} + \dfrac{2^2}{1+x^{2^2}}+ \cdots+ \dfrac{2^n}{1+x^{2^n}}\bigg)\Bigg) \\ &\text{NOTE} :\dfrac{2^k}{1-x^{2^k}}+\dfrac{2^k}{1+x^{2^k}} = \dfrac{2^{k+1}}{1-x^{2^{k+1}}} \\ &=\large \sum_{x=2}^\infty (-1)^x \Bigg( \lim_{n\to\infty}\bigg( -\dfrac{1}{1-x}+\bigg(\dfrac{2^1}{1-x^{2^1}}+ \dfrac{2^1}{1+x^{2^1}}\bigg) + \dfrac{2^2}{1+x^{2^2}}+ \cdots+ \dfrac{2^n}{1+x^{2^n}}\bigg)\Bigg) \\ &=\large \sum_{x=2}^\infty (-1)^x \Bigg( \lim_{n\to\infty}\bigg( -\dfrac{1}{1-x} +\dfrac{2^{n+1}}{1+x^{2^{n+1}}}\bigg)\Bigg) \\ &=\large \sum_{x=2}^\infty \dfrac{(-1)^x}{x-1} \\ &= 1 + \dfrac{-1}{2} + \dfrac{1}{3} + \cdots \cdots \\ &=\int_{0}^{1}\bigg(\large \sum_{n=0}^\infty (-1)^n t^n \bigg)dt \\ &\text{Apply Sum of infinite GP} \\ &=\int_{0}^{1}\dfrac{1}{1+t} dt \\ &=ln(2) \end{aligned}$